# Intensity

## Practice

### practice problem 1

- in the air on the surface of the Earth
- in the ocean near the equator, 1000 m below sea level (in the equatorial SOFAR channel)

#### solution

Sound is a variation in pressure that ranges from…

*P* = *P*_{0} − ∆*P*

to…

*P* = *P*_{0} + ∆*P*

where…

P = |
instantaneous pressure |

P_{0} = |
static pressure |

∆P = |
maximum dynamic pressure |

No sound wave may vary in pressure more than the static pressure of the medium itself. If it did, the minimum pressure would be less than zero, which is impossible. Sound pressure levels are computed relative to an agreed upon standard reference pressure. This number is different for gases and liquids.

For this part, we'll use the standard value of atmospheric pressure (101,325 Pa) and the reference pressure in air (20 μPa).

*L*= 20 log_{P}⎛

⎜

⎝∆ *P*⎞

⎟

⎠*P*_{0}*L*= 20 log_{P}⎛

⎜

⎝101,325 Pa ⎞

⎟

⎠20 × 10 ^{−6}Pa*L*= 194 dB_{P}Begin by computing the pressure at depth.

*P*=*P*_{0}+ ρ*gh**P*= (101,325 Pa) +(1025 kg/m ^{3})(9.8 m/s^{2})(1000 m)*P*= 10,146,325 PaThis is our maximum dynamic pressure. There's no reason to write all those digits. Let's call it…

Δ

*P*= 10 MPa = 10^{7}PaCompared to the reference pressure in water…

*P*_{0}= 1 μPa = 10^{−6}Pa…that's 13 orders of magnitude greater. We can do the rest of this problem without a calculator.

*L*= 20 log_{P}⎛

⎜

⎝∆ *P*⎞

⎟

⎠*P*_{0}*L*= 20 log_{P}⎛

⎜

⎝10 ^{7}Pa⎞

⎟

⎠10 ^{−6}Pa*L*= (20 × 13) dB_{P}*L*= 260 dB_{P}An interesting side note. Negative pressures are possible in liquids. Imagine a liquid in a cylinder with a tightly fitting piston on one side. Push down on the piston and the pressure increases. Pull up and the pressure decreases. Pull up with a force greater than the weight of the liquid and the pressure can be made to drop below zero. The liquid is under tension — a state that is said to be metastable, meaning it's difficult to maintain. At some point, the metastable liquid will "break up" and bubbles of water vapor will form in a process known as cavitation. The water would essentially boil without heat being added. The largest tension reached in water was −140 MPa at 42 °C in 1991 by a group at Arizona State University. (The container was a microscopic cavity in a quartz crystal.) Preventing water from cavitating while being stretched is very difficult. No doubt any sound capable of producing negative pressures in water would be so aggressive that cavitation would take place immediately.

### practice problem 2

- pressure amplitude
- density amplitude
- displacement amplitude
- velocity amplitude
- acceleration amplitude

#### solution

The solutions.

For pressure amplitude, compare 60 dB to the reference intensity of 20 μPa at 0 dB. Start with the equation for pressure level in decibels.

*L*= 20 log_{P}⎛

⎜

⎝∆ *P*⎞

⎟

⎠∆ *P*_{0}Solve for ∆

*P*.*L*_{P}20 dB = log ∆ *P*∆ *P*_{0}10 ^{LP/20 dB}= ∆ *P*∆ *P*_{0}∆ *P*= ∆ *P*_{0}10^{L/20 dB}Substitute and compute (a.k.a. plug and chug).

∆ *P*= (20 μPa)10^{60 dB/20 dB}

∆*P*= (20 μPa)10^{3}

∆*P*= 20 mPaJust a three place decimal shift written as a prefix change from micro to milli.

We have an equation that relates density amplitude to pressure amplitude.

∆ρ = ∆ *P**v*^{2}Let's assume the conversation is taking place in air at room temperature (20 °C) with a speed of sound of 343 m/s. Divide and conquer!

∆ρ = (0.020 Pa) (343 m/s) ^{2}∆ρ = 2.5 × 10 ^{−7}kg/m^{3}I can't relate to numbers this small.

Before we can compute the displacement amplitude, we'll need the intensity. The density of air at room temperature (20 °C) is 1.207 kg/m

^{3}.*I*=∆ *P*^{2}2ρ *v**I*=(0.020 Pa) ^{2}2(343 m/s)(1.207 kg/m ^{3})*I*=4.8 × 10 ^{−7}W/m^{2}Intensity allows us to find displacement amplitude.

*I*=2π ^{2}ρ*f*^{2}*v*∆*s*^{2}∆ *s*=√ *I*2π ^{2}ρ*f*^{2}*v*Human speech doesn't occur at just one frequency. Early 20th century telephone engineers determined that most of the power in human speech fit between 300 and 3,400 Hz. Let's take the geometric mean of this range…

√(300 Hz × 3400 Hz) ≈ 1000 Hz

and continue.

∆ *s*=√ 4.8 × 10 ^{−7}W/m^{2}2π ^{2}(1.207 kg/m^{3})(1000 Hz)^{2}(343 m/s)∆ *s*= 7.7 × 10^{−9}m = 7.7 nmA nanometer is about ten atoms in a row. This means the air molecules driving your eardrum only wiggle back and forth a distance of about 150 atoms when listening to human speech. (I doubled the displacement amplitude since they are moving that distance on both sides of their equilibrium position.)

Velocity amplitude is most easily computed from displacement amplitude.

∆ *v*= 2π*f*∆*x*

∆*v*= 2π(1000 Hz)(7.7 × 10^{−9}m)

∆*v*= 4.8 × 10^{−5}m/sAnother small number that means nothing to me.

For acceleration amplitude.

∆ *a*= 2π*f*∆*v*

∆*a*= 2π(1000 Hz)(4.8 × 10^{−5}m/s)

∆*a*= 0.30 m/s^{2}Not such a small number anymore. Not a lot when compared to gravity, but not microscopic like the other amplitudes.

### practice problem 3

- How does the intensity of the sound wave in air compare to the same sound wave in water?
- By how many decibels is the sound intensity level reduced?
- Where does the excess energy go when a sound wave is incident upon an air-water interface?

air | water | |
---|---|---|

density (kg/m^{3}) |
1.225 | 0999 |

speed of sound (m/s) | 0,340 | 1497 |

#### solution

The intensity of a sound wave depends not only on the pressure of the wave, but also on the density of the medium and speed of sound in the medium.

*I*=∆ *P*^{2}2ρ *v*Compare the intensity of the sound wave in the two media.

*I*_{air}= ∆ *P*^{2}/2ρ_{air}v_{air}*I*_{water}∆ *P*^{2}/2ρ_{water}v_{water}When a sound wave travels from air to water, pressure is assumed to be the property that is transferred from one medium to the other. Cancel out the constant terms and simplify the compound fraction.

*I*_{air}= ρ _{water}v_{water}*I*_{water}ρ _{air}v_{air}Insert numbers…

*I*_{air}= (999 kg/ ^{3})(1497 m/s)*I*_{water}(1.225 kg/ ^{3})(340 m/s)and compute.

*I*_{air}= 3590 *I*_{water}Water is about 800 times more dense than air and has a speed of sound 4.5 times faster. Higher density and higher sound speed both give a lower intensity. Sounds with the same pressure amplitude are about 3600 times more intense in air than in water.

Use the equation for sound intensity level in decibels.

*L*= 10 log_{I}⎛

⎜

⎝*I*⎞

⎟

⎠*I*_{0}Let the intensity in the air be the reference value.

*L*= 10 log_{I}⎛

⎜

⎝*I*_{water}⎞

⎟

⎠*I*_{air}Then the level drop is…

*L*= 10 log_{I}⎛

⎜

⎝1 ⎞

⎟

⎠3590 *L*= −36 dB_{I}Intensity is a measure of power per area and power is the rate at which energy is transferred, so intensity is a surrogate for energy. Very little of the energy of a sound made above the surface of the water water is transmitted into the water. Energy cannot be created or destroyed, so it has to go somewhere. The excess energy is reflected off the air-water interface. We call that an echo.

### practice problem 4

#### solution

Answer it.